Problem: There are 12 different-colored crayons in a box. How many ways can Karl select four crayons if the order in which he draws them out does not matter?
Answer: There are 12 ways to select the first crayon, 11 ways to select the second, 10 ways to select the third, and 9 ways to select the last. However, since order does not matter, we must divide by the number of ways he can draw out the crayons, which is $4!$.

The answer is $\dfrac{12\times11\times10\times9}{4!}=\boxed{495}$ ways.